Dorian Popa Constantin-Cosmin Todea

SPECIAL MATHEMATICS. PROBLEMS

U.T. PRESS

CLUJ-NAPOCA

2014

ISBN 978-973-662-988-4

Editura U.T.PRESS

Str.Observatorului nr. 34

C.P.42, O.P. 2, 400775 Cluj-Napoca

Tel.:0264-401.999 / Fax: 0264-430.408

e-mail: utpress@biblio.utcluj.ro

www.utcluj.ro/editura

Director: Prof. dr. Daniela Manea

Consilier editorial: ing. Calin D. Campean

Copyright c2014 Editura U.T.PRESSReproducerea integrala sau partiala a textului sau ilustratiilor din aceasta carte

este posibila numai cu acordul prealabil scris al editurii U.T.PRESS.

Multiplicarea executata la editura U.T.PRESS.

ISBN978-973-662-988-4

Contents

Preface 1

1 Differential equations effectively integrable 2

1.1 Differential equations with separable variables . . . . . . . . . . . . . 2

1.2 Homogenous equations . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Linear equations of order one . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Bernoullis equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.5 Riccatis equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6 Exact differential equations. Integrant factor . . . . . . . . . . . . . . 10

1.7 Equations of Clairaut and Lagrange . . . . . . . . . . . . . . . . . . . 13

1.8 Higher order differential equations . . . . . . . . . . . . . . . . . . . . 15

1.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Linear differential equations of order higher than one 20

2.1 Linear differential equations with constant coefficients . . . . . . . . . 20

2.2 Eulers equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Systems of differential equations and partial differential equations 26

3.1 Linear systems with constant coefficients . . . . . . . . . . . . . . . . 26

3.2 Symmetric Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.3 Partial differential equations . . . . . . . . . . . . . . . . . . . . . . . 34

3.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4 Complex Analysis 43

4.1 Complex numbers. Basic results . . . . . . . . . . . . . . . . . . . . 43

4.2 Complex functions of a complex variable . . . . . . . . . . . . . . . . 46

4.3 Laplaces transform. . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

i

ii

Bibliography 62

Preface

The aim of this book is to cover the analytical program for the course of Special Math-

ematics of different sections or faculties from Technical University of Cluj-Napoca. It

is however, mainly addressed for the students of Electrical Engineering Faculty, which

follow this course in the second semester of the first year. The book has four chapters

and each chapters ends with a section of proposed problems. The first three chapters

are dedicated to differential equations and in the fourth chapter we introduce the

reader into the basics of complex analysis. One of the main features of this book is,

in the opinion of the authors, the multitude of detailed solved problems which come

to help the students. We used as sources for selecting some proposed and solved

problems the following references: [1], [2], [3].

Cluj-Napoca, 2014

The Authors

1

Chapter 1

Differential equations effectivelyintegrable

In this chapter we present the most relevant types of differential equations of order

one and some basic and elementary techniques techniques to solve them. We end this

chapter with a section regarding some higher order differential equations.

1.1 Differential equations with separable variables

A differential equation of the form

y = f(x)g(y) (1.1.1)

where f C(I), g C(J), I, J R are intervals is called equation with separablevariables. For y J1 J , J1 interval with g(y) 6= 0 the equation 1.1.1 is equivalentto

dy

g(y)= f(x)dx

and the solution follows by integrationdy

g(y)=

f(x)dx.

If y0 J and g(y0) = 0 then the equation (1.1.1) admits the singular solution y(x) =y0, for all x I.

Example 1.1.1. Integrate xy(1 + x2)y = 1 + y2.

2

3Proof. We have xy(1 + x2) dydx

= 1 + y2, hence ydy1+y2

= dxx(1+x2)

. We integrate to obtainydy1+y2

=

dxx(1+x2)

. Equivalently

ydy1+y2

= (

1x x

1+x2

)dx, hence

1

2ln(y2 + 1) = ln | x | 1

2ln(1 + x2) + C.

It follows that ln(y2 + 1) = lnx2 ln(1 + x2) + C2. For C2 = ln k, k 1 we obtainthe solution

y2 + 1 =kx2

1 + x2.

Example 1.1.2. Integrate y = (x+ y + 1)2.

Proof. Let x + y + 1 = z, z = z(x). Then y = z x 1 and y = z 1 hence theequation becomes z = 1 + z2, that is dz

1+z2= dx. Integrating the previous relation it

follows that dz

1 + z2=

dx

or

arctan z = x+ C

so z = tan(x+ C) hence the solution

y(x) = tan(x+ C) x 1.

1.2 Homogenous equations

A differential equation of the form

y = f(y

x) (1.2.1)

4where f C(I), with I R an interval and f(u) 6= u for any u I, is calledhomogenous equation.

Let z : I R be the function defined by z(x) = y(x)x, x I. Then y = xz and

y = xz + z hence the equation becomes xz = f(z) z, i.e.dz

f(z) z =dx

x

which is an equation with separable variables.

Example 1.2.1. Integrate 2x3y = y(3x2y + y2), x (0,).

Proof. We have y = 3x2+y3

2x3, that is y = 3

2yx

+ 12( yx)3. The substitution y

x= z, z = z(x)

leads to

xz + z =3

2z +

1

2z3.

Equivalently, to xz = 12z + 1

2z3. The new equation becomes 2dz

z(1+z2)= dx

xand by

integration we get

2

(1

z z

1 + z2

)dz =

dx

x

hence

ln z2 ln(1 + z2) = ln x+ lnC,C 0.

Finally by replacing z = yx

in z2

1+z2= Cx, we get y2 = Cx

3

1Cx , x (0,).

Example 1.2.2. Prove that the differential equation

y = f(a1x+ b1y + c1a2x+ b2y + c2

)f C(I), ak, bk, ck R, k {1, 2} becomes an homogenous equation if the system ofequations a1x+ b1y + c1 = 0a2x+ b2y + c2 = 0admits a unique solution (x0, y0).

5Proof. Consider the change of variables

x = t+ x0y = u+ y0 , and let u = u(t) be theunknown function. Then

a1x+ b1y + c1 = a1(t+ x0) + b1(u+ y0) + c1 = a1t+ b1u,

a2x+ b2y + c2 = a2(t+ x0) + b2(u+ y0) + c2 = a2t+ b2u

and y = dydx

= dudt. Therefor we obtain

du

dt= f

(a1t+ b1u

a2 + b2u

)= f

(a1 + b1

ut

a2 + b2ut

)= g

(ut

),

which is an homogenous equation.

1.3 Linear equations of order one

A differential equation of the form

y + f(x)y = g(x) (1.3.1)

where f, g C(I), I R is an interval is called a linear differential equation of orderone.

Let F (x) = xx0f(t)dt, x0 I be an antiderivative of f . Multiplying the equation

(1.3.1) by eF (x) we get

y(x) = eF (x)( x

x0

g(t)eF (t)dt

), C R.

The function h : I R is called the integrating factor of the above equation.

Example 1.3.1. Integrate y + 2y = ex, x R.

Proof. We compute F (x) = e2x, x R, so we multiply with e2x and the equationbecomes y(x)e2x + 2e2xy(x) = ex or (y(x)e2x) = ex , hence by integration we obtain

y(x) = e2x(x+ C), C R.

6Example 1.3.2. Let f : [0,) R be a continuous function such that there existslimx

f(x) = l, l R and a > 0. Prove that any solution of the equation

y + ay = f(x)

admits an horizontal asymptote at +.

Proof. Let y be a solution of the equation. Multiplying with eax the equation becomes

(y(x)eax) = f(x)eax, x [0,). Integrating on [0, x], x > 0 it follows

y(x)eax =

x0

f(t)dt+ C,C R

hence

y(x) =

x0f(t)eatdt+ C

eax, x > 0.

Then

limx

y(x) = limx

( x0f(t)eatdt

eax+

C

eax

)= lim

x

( x0f(teatdt)

)aeax

=

limx

f(x)eax

aeax=l

a.

Thus y = la

is the horizontal asymptote of f at .

Example 1.3.3. Find all continuous function y : R R satisfying the followingintegral equation

x0

(x s)y(s)ds = x0

y(s)ds+ sinx, x R.

Proof. Since y is a continuous function, the functions from the left and the right hand

are differentiable. First we put the equation in the form

x

x0

y(s)ds x0

sy(s)ds =

x0

y(s)ds+ sinx

7and by differentiation with respect to x we get

xy(x) +

x0

y(s)ds xy(x) = y(x) + cos x (?)

A second differentiation leads to y(x) = y(x) sinx, a linear equation of order one,which can be written in the form (y(x)ex) = ex sinx. By integration we get

y(x) = Cex sinx+ cosx2

, x R.

Take x = 0 in relation (?) to obtain y(0) = 1 so y(0) = C 12

= 1, hence C = 12.

The solution is

y(x) = ex + sinx+ cosx

2, x R.

1.4 Bernoullis equations

A differential equation of the form

y + f(x)y = g(x)y, R \ {0, 1}

where f, g C(I), I R interval, is called Bernoullis equation. For > 0 theequation admits the solution y(x) = 0, x I. On an interval I1 I where y(x) 6=0, x I1 the substitution z(x) = y1(x), x I leads to

z + (1 )f(x)z = (1 )g(x)

which is a linear equation.

Example 1.4.1. Integrate the equation xy2y = x2 + y3.

8Proof. We can divide with y2 to obtain the rigorous form of this Bernoulli equation,

or directly we make the substitution z = y3 and notice that z = 3y2y. By replacing

these in the equation we obtain

x

3z = x2 + z,

next we multiply by 3x

to obtain the linear equation z 3xz = 3x. We multiply it by x3

and we have (x3z) = 3x2, hence z = 3x3x2dx. Equivalently z = 3x3(C x1).

Finally we obtain y3 = Cx3 3x2.

Example 1.4.2. Integrate the equation y + xy = xy2, x R.

Proof. We use the substitution z(x) = y1(x), then z(x) = y2(x)y(x). We multi-ply the given equation by y2 and we obtain

y2y + xy1 = x.

From the above substitutions the linear equation in z is z + xz = x, that is

z xz = x.

Since the antiderivative is F (x) = x22

, we multiply this equation by ex2

2 to obtain

ex2

2 z xex2

2 z = xex2

2 .

Equivalently we have (ex2

2 z) = xex2

2 , thus ex2

2 z =xe

x2

2 dx. Integrating we

have

z = ex2

2 (ex2

2 C).

We recall that z = y1 to conclude that

y1(x) = ex2

2 (ex2

2 C).

91.5 Riccatis equations

A differential equation of the form

y = f(x)y2 + g(x)y + h(x)

where f, g, h C(I), I R interval, is called Riccatis equation. Generally Riccatisequations cannot be effectively integrated. But if y0 is a particular solution of it, then

the substitution y = y0 +1z

gives the linear differential equation

z (2f(x)y0(x) + g(x))z = f(x).

Example 1.5.1. Integrate the equation y = y2 1xy 1

x2, if it admits the particular

solution y0(x) = 1x .

Proof. The substitution y = 1x

+ 1z

leads to the equation

(1x

+1

z) = (1

x+

1

z)2 1

x(1x

+1

z) 1

x2,

equivalent to

1

x2 z

z2=

1

x2 2xz

+1

z2+

1

x2 1xz 1x2.

We obtain the linear equation in z:

z

z2= 3

xz+

1

z2,

next we multiply by xz2 to obtain xz = 3z+x or z 3xz = 1. We multiply this

equation by x3 to obtain zx3 3x4 = x3, that is (zx3) = x3. Integratingwe have z = x3 x3dx, hence z(x) = x3(x2

2+ C), with the final solution

y = 1x

+1

x3(x22

+ C), C R.

10

Example 1.5.2. Integrate (1 + x3)y y2 x2y 2x = 0, x > 1, if the equationadmits a particular solution of the form y0(x) = ax

n, a R, n N.

Proof. Replacing y0 in the equation we get

(a+ x3)naxn1 a2x2n axn+2 2x = 0

or

a2x2n + (na a)xn+2 + na2xn1 2x = 0, x > 1.

It follows n = 2 and a2x4 +ax4 + 2ax 2x = 0 for all x > 1, hence a = 1. Theparticular solution is y0(x) = x

2. The substitution is y = x2 + 1z

and leads to

z(1 + x3) + 3x2z = 1 or (z(1 + x3)) = 1.

So z(1 + x3) = x+ C hence z = x+C1+x3

. Finally

y = x2 +1 + x2

x+ C =Cx2 + 1

C x , C R.

1.6 Exact differential equations. Integrant factor

Let D R be a rectangle and P,Q C1(D). A differential equation of the form

P (x, y)dx+Q(x, y)dy = 0 (1.6.1)

where Py

(x, y) = Qx

(x, y) for all (x, y) D is called an exact differential equation.Under the previous conditions there exists a function F C2(D) given by the relation

F (x, y) =

xx0

P (t, y)dt+

yy0

Q(x0, t)dt, (x0, y0) D,

11

such that dF (x, y) = P (x, y)dx + Q(x, y)dy, (x, y) D. Since the exact differentialequation is equivalent to dF (x, y) = 0 the solutions are implicitly defined by

F (x, y) = C,C R.

The function F is called an antiderivative (primitive) of the differential form Pdx +

Qdy.

If an equation of the form 1.6.1 is not an exact equation then a function C1(D)with the property that the equation

(x, y)P (x, y)dx+ (x, y)Q(x, y)dy = 0 (1.6.2)

is an exact differential equation is called integrant factor. Denoting P1(x, y) =

(x, y)P (x, y), Q1(x, y) = (x, y)Q(x, y), (x, y) D the equation 1.6.2 is an ex-act equation if

P1(x, y)

y=Q1(x, y)

x, (x, y) D (1.6.3)

The equation 1.6.3 is equivalent to the equation of integrant factor

Q

x P

y=

(P

y Qx

). (1.6.4)

In practice, usually we are looking for integrant factors of the form = (x), =

(y). If the equation

Q(x) =(P

y Qx

)(x)

depends only on x. then there exists = (x). If the equation

P(y) =(P

y Qx

)(y)

depends only on y, then there exists = (y).

Example 1.6.1. Integrate the equation eydx (2y + xey)dy = 0, x, y R.

Proof. It is easy to check that P (x,y)y

= Q(x,y)x

= ey where

P (x, y) = ey, Q(x, y) = 2y xey.

12

We apply the integral formula for F to obtain

F (x, y) =

xx0

eydt+ yy0

(2t x0et)dt = eyt|xx0 2[t2

2

]yy0

x0[ey]y

y0=

ey(x x0) (y2 y20) + x0(ey ey0) = xey x0ey y2 + y20 + x0ey x0ey0 =

xey y2 + y20 x0ey0 .

Since x0, y0 are constants the solution of the equation is xey y2 = C.

Example 1.6.2. Find the integrant factor = (y), depending on y, for the equation

(2xy2 3y3)dx+ (7 3xy2)dy = 0

Proof. We have

P (x, y)

y= 4xy 9y2, Q(x, y)

x= 3y2.

Since depends on y, = (y) we apply the above formula to obtain

(2xy2 3y3)(y) = (4xy 9y2 + 3y2)(y)

We divide by y to obtain y(2x 3y)(y) = 2(2x 3y)(y). Next we divide by2x 3y to obtain an equation with separable variables y d

dy= 2 hence d

= 2dy

y.

Integrating we have ln | |= 2 ln | y | +C, thus (y) = Cy2.

Example 1.6.3. Find the integrating factor = (x+ y2), depending on x+ y2 for

the equation (3y2 x)dx+ (2y3 6xy)dy = 0.

Proof. We have P (x,y)y

= 6y, Q(x,y)x

= 6y. For shortness we denote by t theexpression x+ y2, hence t = t(x, y) = x+ y2. We apply formula 1.6.3 to obtain

Q(t)t

x P(t) t

y=

(P

y Qx

)(t).

13

Since tx

= 2x and ty

= 1 we obtain

(t)[(2y3 6xy)1 (3y2 x)2y] = [6y (6y)](t),

equivalently (t)(2y3 6xy 6y3 + 2xy) = 12y(t). Divide by y and we get(t)(4y2 4x) = 12(t). Divide also by 4 and we obtain the equation

(x+ y2)(t) = 3(t)

that is tddt

= 3 which is a separable variable equation, henced

= 3dt

t.

We integrate and obtain ln | |= 3 ln | t | + ln | C |, thus (t) = Ct3. Weconclude that an integrant factor can be chosen (x, y) = (x+ y2)3.

1.7 Equations of Clairaut and Lagrange

A differential equation of the form

y = xy + g(y))

where g C(I), I R interval, is called Clairauts equation. We make the sub-stitution y = p to obtain the equation y = xp + g(p). By differentiating we getpdx = pdx + dxp + g(p)dp hence [x + g(p)]dp = 0. We obtain the singular solution(given by parametric equations )

x = g(p), y = pg(p) + g(p)

and from dp = 0 we obtain the general solution (p = C)

y = Cx+ g(C).

A differential equation of the form

y = xf(y) + g(y))

14

where f, g C(I), I R interval, with f(y) 6= y , is called Lagranges equation. Thesame substitution y = p leads us to y = xf(p) + g(p) hence, by differentiating we getdy = dxf(p) + xf (p)dp+ g(p)dp. We obtain the linear equation

(p f(p))dx = (xf (p) g(p))dp,

with the unknown x = x(p), which has a general solution x = h(p, C). WE obtain

the general parametric solution of the Lagranges equation

x = h(p, C), y = h(p, C)f(p) + g(p).

Example 1.7.1. Integrate the equation y = xy +

1 + y2.

Proof. Let y = p, so y = xp+

1 + p2. Differentiate to obtain

dy = dxp+ xdp+2p

2

1 + p2dp

hence

pdx = pdx+ xdp+p

1 + p2dp

First we have dp = 0 that is p = C and the general solution y = xC +

1 + C2.

Secondly we obtain the singular solution

x = p1 + p2

, y = p2

1 + p2+

1 + p2.

Example 1.7.2. Integrate the equation y = x(1 + y) + y2.

Proof. Let y = p then y = x(1 + p) + p2 and we differentiate to obtain dy = dx(1 +

p) + xdp+ 2pdp, hence pdx = dx+ pdx+ xdp+ 2pdp that is

dx = (x+ 2p)dp

15

which is a linear equation with the unknown x = x(p).We write this equation with

derivative dxdp

= x+ 2p, that is x x = 2p, hence x + x = 2p. We multiply thisequation by ep to obtain epx+ epx = 2pep, that is (epx) = 2pep. Equivalently weget

x = 2eppepdp = 2ep(pep ep + C) = 2p+ 2 2Cep.

We replace x in the first formula of y to obtain

y = (1 + p)(2p+ 2 2Cep) + p2.

So, the parametric equations are

x = 2p+ 2 2Cep, y = (1 + p)(2p+ 2 2Cep) + p2.

1.8 Higher order differential equations

We present some classes of differential equations of order n, with n a positive integer

greater than one, which can be reduced to differential equations of order strictly less

than n.

1. Equations of the form F (x, y(k),y(k+1)

, . . . , y(n)) = 0 with the unknown

y Cn(I), y = y(x), I R interval. The substitution z = y(k) leads to the equationof order n k

F (x, z, z, . . . , z(nk)) = 0.

Example 1.8.1. Integrate y + 2y = e2x, x R.

Proof. The substitution z = y leads to the equation z + 2z = e2x. Multiplying by

e2x we get (ze2x) = 1 so z(x) = e2x(x+ C1) and

y(x) =

e2x(x+ C1)dx = 1

2e2x(x+ C1 +

1

2) + C2, C1, C2 R.

16

2. Equations of the form F (y, y, y, . . . , y(n)) = 0. Let y = p(y) where pbecomes the new unknown of the equation. We have

y = p;

y = p(y)y = pp;

y = p(y)(y)2 + p(y)y = p2p + p(p)2;

. . .

so we get a differential equation of order (n 1) with the unknown p.

Example 1.8.2. Integrate 2yy = (y)2 + 1.

Proof. The substitution y = p, p = p(y) leads to 2ypp = p2 + 1 which can be written

as 2pdpp2+1

= dyy

. It follows

2pdpp2+1

=

dyy

, hence ln(p2 + 1) = ln | Cy | so p = Cy 1.To obtain the solution we have to integrate the equation y = Cy 1, which isequivalent to dy

Cy1 = dx. We get 4(Cy 1) = C2(x+ C1), C, C1 R.

3. Equations of the form F (x, yy, yy, . . . , y

(n)

y) = 0. Remark that these equa-

tions are homogenous with respect to y, y, . . . , y(n). The substitution

z =y

y, z = z(x)

leads to a differential equation of order n 1. Indeedy

y= z2 + z;

y

y= z3 + 3zz + z, . . .

Example 1.8.3. Integrate the equation x2yy = (y xy)2.

Proof. Divide by y2 to obtain x2 yy

= (1 xyy

)2. Take the substitution yy

= z, hence

yy

= z + z2 and replace it to obtain the equation x2(z + z2) = (1 xz)2, that is

x2z + x2z2 = 1 + x2z2 2xz.

17

We have a linear equation x2z + 2xz = 1 which is equivalent to (x2z) = 1, hence

x2z = x+ C. Since z = yy

we obtain the separable variable equation

y

y=x+ C

x2.

Equivalently we get dyy

= ( Cx2

+ 1x)dx. Integrating we obtain ln | y |= C 1

x+ ln | x |

+C1 hence the solution

y = xeCxC1.

1.9 Problems

Problem 1.9.1. Integrate the following differential equations of order one:

(1) xydx+ (x+ 1)dy = 0;

(2)y2 + 1dx = xydy;

(3) 2x2yy + y2 = 2;

(4) (x y)dx+ (x+ y)dy = 0;

(5) xy y = x tan yx;

(6) x y 1 + (y x 2)y = 0;

(7) xy 2y = 2x4;

(8) y + y tanx = 1cosx

;

(9) y + 2y = y2ex;

18

(10) xy y2 + (2x+ 1)y = x2 + 2x, if admits the particular solution y0 = x;

(11) y = y2 2yex + e2x + ex, if admits the particular solution y0 = ex;

(12) y + y2 2y sinx+ sin2 x = cosx, if admits the particular solution y0 = sinx;

(13) y = xy ln y;

(14) y = xy + y2;

(15) y = 2xy + sin y;

(16) y = 2xy + ln y;

Problem 1.9.2. Integrate the exact differential equations:

(1) (2 9xy2)xdx+ (4y2 6x3)ydy = 0;

(2) yxdx+ (y3 + lnx)dy = 0;

(3) 3x2+y2

y2dx 2x3+5y

y3dy = 0;

(4) ydx+ xdy = 0;

Problem 1.9.3. Find the integrant factor for the equations:

(1) (x+ sinx+ sin y)dx+ cosydy = 0 depending on x, = (x);

(2) (x y)dx+ (y + x2)dy = 0 depending on x2 + y2, = (x2 + y2);

Problem 1.9.4. Integrate the following equations of order higher than one:

(1) x2y = y2;

(2) xy = y + x2;

19

(3) y2 + 2yy = 0;

(4) y3y = 1;

(5) yy y2 = y2y;

(6) x2yy = (y xy)2;

(7) y + 2xyy = 0;

(8) xyy xy2 = yy.

Solutions: 1.9.1 (1). y = C(x+ 1)ex; (2). ln | x |= C +y2 + 1; (3). y2 2 =Ce

1x ; (4). ln(x2 + y2) = C 2 arctan y

x; (5). Divide by x to obtain y y

x= tan y

x

and then take the substitution yx

= z to obtain a variable separable equation with

the final solution sin yx

= Cx; (6). Make the substitution z = y x, z = z(x)and obtain the final solution (y x + 2)2 + 2x = C; (7). y = Cx2 + x4; (8).y = sin x + C cosx; (9). y = 0 and y(ex + Ce2x) = 1; (10). Take the substitution

y = z + x which leads to the Bernoulli equation in z,xz z2 + z = 0 and the finalsolution y = 1

Cx+1+x; (11). take the substitution y = z+ex and get the final solution

y = 1Cx + e

x; (12). y = 1C+x

+ sinx; (13). y = Cx lnC and y = 1 + lnx; (14).y = Cx + C2 and x2 + 4y = 0; (15). x = Cp sin pcos p

p2, y = 2Cp sin p2 cos p

p; (16).

x = Cpp2, y = 2(Cp)p

p+ ln p; 1.9.2 (1). x2 3x3y2 + y4 = C; (2). 4y lnx+ y4 = C;

(3). x + x3

y2+ 5

y= C; (4). xy = C; 1.9.3 (1). = ex; (2). = (x2 + y2)

32 ; 1.9.4 (1).

The substitution is y = z and the final solution

C1x C21y = ln | C1x+ 1 | +C 2;

(2). y = x3

3+ C1

x2

2+ C2; (3). y

= p, y = pp and the final solution is y = C andy3 = C1(x + C2)

2; (4). C1y2 1 = (C1x + C2)2; (5). The equation in p = p(y) is

p 1yp = y with the solution y = C1 and

yy+C2

= C3eCx; (6). Divide by y2 to obtain

the equation x2 yy

= (1 xyy

). Use the substitution yy

= z to obtain the equation

z = 2zx

+ 1x2

The final solution is y = xeC1x C2; (7). Divide by y

2; (8). y = C2eC1x2 ,

Chapter 2

Linear differential equations oforder higher than one

2.1 Linear differential equations with constant co-

efficients

To integrate a homogenous linear differential equation with constant (real) coefficients

a0y(n) + a1y

(n1) + . . .+ an1y+ any = 0

we attache the (polynomial) characteristic equation

a0rn + a1r

n1 + . . .+ an1 + an = 0

and we find all its roots r1, . . . , rn C. A root r from this set may be a multiple realroot of order k with k 1 or it may be a multiple complex root and its conjugater = + i, r = i, 6= 0 of order k with k 1. The general solution of theabove differential equation is the sum of all terms (associated to the roots) of form

(C1 + C2x+ . . .+ Ckxk1)erx

if r R is a multiple root of order k and of formPk1(x)ex cos x+Qk1(x)ex sin x

if + i and i are roots of order k. Here Pk1, Qk1 are polynomials of degreek 1 with coefficients arbitrary constants.

20

21

Example 2.1.1. Integrate the equation y(5) 2y(4) 16y + 32y = 0.Solution. The characteristic equation is r5 2r4 16r+ 32 = 0 hence, by decom-

posing we have

r4(r 2) 16(r 2) = 0(r 2)(r 2)(r + 2)(r2 + 4) = 0.

The roots are r1 = r2 = 2 multiple of order 2; r3 = 2 simple of order one; r4 =2i, r5 = 2i simple of order one. We apply the above regulae to obtain the generalsolution

y = (C1 + C2x)e2x + C3e

2x + C4 cos 2x+ C5 sin 2x.

To integrate a non-homogenous linear differential equation with constant (real)

coefficients

a0y(n) + a1y

(n1) + . . .+ an1y+ any = f(x), f(x) 6= 0

where f(x) is sums and products of functions like: b0+b1x+. . .+bmxm, ex, cos x, sin x

we solve first the homogenous equation and find its general solution yo and then we

search for a particular solution yp with the method of nondeterminate coefficients.The

general solution will be

y = yo + yp.

If f(x) is of the form

ex(P (x) cos x+Q(x) sin x)

we search a particular solution of the form

yp = xsex(Rm(x) cos x+ Tm(x) sin x),

where s is equal to zero if +i is not a root of the characteristic equation and is the

order of multiplicity of the root + i otherwise. Rm, Tm are polynomials of degrees

greater or equal to the degrees of P and Q. For finding the coefficients of Rm, Tm we

replace yp in the non-homogenous equation and we identify the same terms.

Example 2.1.2. Integrate the equation y 6y + 9y = xe3x + e3x cos 2x.Solution. The characteristic equation is r3 6r2 + 9r = 0 and has r1 = r2 = 3 a

root of order 2 and r3 = 0 a simple root. So, the general solution of the homogenous

equation is

yo = (C1 + C2x)e3x + C3.

22

The second member of the equation has two terms of different forms: for the first

xe3x, we have = + i = 3 and for the second e3x cos 2x, we have + i = 3 + 2i.

These numbers are different hence we search two separate particular solutions of the

equations

y 6y + 9y = xe3x, (1)y 6y + 9y = e3x cos 2x (2).

The number = 3 is a root of order 2 so a particular solution for (1) is of the form

yp1 = x2(ax + b)e3x. We replace it in (1) to obtain a = 1

18, b = 1

18. Moreover

+ i = 3 + 2i is not a root for the characteristic equation hence we search for a

particular solution of the equation (2) of the form yp2 = e3x(c cos 2x + d sin 2x). We

replace it in (2) to obtain c = 352, d = 1

26. The general solution is

y = yo+yp1 +yp2 = (C1+C2x)e3x+C3+x

2(1

18x 1

18)e3x+e3x( 3

52cos 2x 1

26sin 2x).

Example 2.1.3. Integrate the equation y + 4y = sinx sin 2x.

Proof. The characteristic equation is r2 + 4 = 0 with the complex conjugate

roos of order one r1 = 2i, r2 = 2i. So, the solution of the homogenous equation isyo = C1 cos 2x+ C2 sin 2x.

For the non-homogenous part f(x) = sinx sin 2x we use the formula sin a sin b =12(cos(a b) cos(a + b)) to obtain f(x) = 1

2cosx 1

2cos 3x. The first term 1

2cosx

give the number = 1 which is not a root of the characteristic equation and the

second term 12

cos 3x gives a different number = 3 which again is not a root of the

characteristic equation. Hence we will search for two separate particular solutions of

the equations

y + 4y =1

2cos 2x, (1)

y + 4y = 12

cos 3x (2).

A particular solution is of the form yp1 = a cosx with yp1

= a sinx and yp2 =a cosx. We replace these terms in (1) to obtain

a cosx+ 4a cosx = 12

cos 2x,

thus 3a = 12

which gives a = 16. The other particular solution for (2) is of the form

yp2 = b cos 3x and if we replace it we obtain b =110

. The general solution is

y = C1 cos 2x+ C2 sin 2x+1

6cosx+

1

10cos 3x.

23

2.2 Eulers equations

The equation of Euler is

a0xny(n) + a1x

n1y(n1) + . . .+ an1xy + any = f(x).

To solve this equation we reduce it to a linear equation with constant coefficients by

using the substitution x = et for x > 0 (or x = et for x < 0). The new linearequation has the attached characteristic equation of the form

a0r(r 1)(r 2) . . . (r n+ 1) + . . .+ an2r(r 1) + an1r + an = 0.

To write the linear equation we change each term xky(k) with a product of k factors

decreasing by unity; r(r 1) . . . (r k + 1). Alternatively, in practice we prefer tochange in the given equation: x by et; xy(x) by y(t); x2y(x) by y(t)y(t); x3y(x)by y(t) 3y(t) + 2y(t) + 2 and so on.

Example 2.2.1. Integrate the equation x3y x2y + 2xy 2y = x3

Solution.. The characteristic equation is

r(r 1)(r 2) r(r 1) + 2r 2 = 0

hence by decomposition we obtain (r 1)(r2 3r + 2) = 0 with the multiple rootr1 = r2 = 1 of order 2 and the simple root r3 = 2. The general solution of the

homogenous linear equation is

yo = (C1 + C2t)et + C3e

2t.

To solve the given Eulers equation we multiply all factors in the characteristic equa-

tion an we obtain r3 4r2 + 5r 2 = 0. Now, immediately we obtain from thisthe linear non-homogenous equation with constant coefficients (obtained with the

substitution x = et > 0):

y 4y + 5y 2y = e3t.Since 3 is not a root of the characteristic equation we search for a particular solution

of the form yp = ae3t. Replacing it we obtain

27ae3t 36ae3t + 15ae3t 2ae3t = e3t

24

hence 4ae3t = e3t. We obtain a = 14

and the general solution is

y = yo + yp = (C1 + C2t)et + C3e

2t +1

4e3t = (C1 + C2 lnx) + C3x

2 +1

4x3, x > 0.

For x < 0 the formula is similar, so the general solution will be

y = (C1 + C2 ln |x|) + C3x2 + 14x3.

Example 2.2.2. Integrate the equation x2y xy 3y = 8x3.

Solution. The characteristic equation is r(r1) r3 = 0 that is r22r3 = 0which has two roots r1 = 3 and r2 = 1. The linear equation is

y 2y 3y = 8e3t

with the homogenous solution yo = C1e3t+C2e

t. Since f(t) = 8e3t and 3 is a root ofthe characteristic equation we search for a particular solution of the form yp = Ate

3t.

We have yp = Ae3t(3t + 1) and yp = Ae

3t(9t + 6). We replace these expressions to

obtain

Ae3t(9t+ 6) 2Ae3t(3t+ 1) 3Ae3tt = 8e3t.It follows

Ae3t(9t+ 6 6t 2 3t) = 8e3t,hence 4A = 8 and A = 2. We obtain

y = yo + yp = C1e3t + C2e

t + 2te3t = C1e3 lnx + C2e lnx + 2 lnx e3 lnx.

The general solution is

y = C1x3 +

C2x

+ 2x3 ln |x|.

2.3 Problems

Problem 2.3.1. Integrate the following differential equations:

(1) y + y 2y = 0;

25

(2) y + 4y + 3y = 0;

(3) y 2y = 0;

(4) y 8y = 0;

(5) y(v) 6y(iv) + 9y = 0;

(6) y + y 2y = 3xex;

(7) y y = 2ex x2;

(8) y 3y + 2y = sinx;

(9) y y + y y = x2 + x

(10) y(iv) + y = 7x 3 cosx.

Problem 2.3.2. Integrate the following Eulers equations:

(1) x2y 4xy + 6y = 0;

(2) x2y xy + y = 8x3;

(3) x2y 3xy + 4y = x+ 2;

(4) x2y 2y = sin ln x;

(5) x2y = 2y;

Solutions: 2.3.1 (1). y = C1ex + C2e

2x; (2). y = C1ex + C2e3x; (3). y = C1 +C2e

2x; (4). y = C1e2x + ex(C2 cos(x

3) +C4 sin(x

3)); (5). y = C1 +C2x+C3x

2 +

e3x(C4+C5x); (6). y = C1ex+C2e

2x+(x2

2 x

3)ex; (7). y = C1e

x+C2ex+xex+x2+2;

(8). y = C1ex+C2e

2x+ 110

sinx+ 310

cosx; (9). y = C1ex+C2 cosx+C3 sinxx2+3x1.

2.3.2 (1). y = C1x2 + C2x

3; (2). y = x(C1 + C2 ln |x|) + C3 ln2 |x|; (3). y =C1x

2 + C2x2 lnx + x + 1

2; (4). y = C1x

2 + C2x1 + 1

10cos lnx 3

10sin lnx; (5). y =

C1 + C2 ln |x|+ C3x3.

Chapter 3

Systems of differential equations

and partial differential equations

3.1 Linear systems with constant coefficients

A linear system with constant coefficients is a system of the form

y1 = a11y1 + a12y2 + . . . a1nyn + f1(x)

y2 = a21y1 + a22y2 + . . . a2nyn + f2(x)

. . .

yn = an1y1 + a1n2y2 + . . . annyn + fn(x)

where y1 = y1(x), . . . yn = yn(x) are the unknown differentiable functions and aij R, fi C1(R), i, j {1, . . . , n}. To solve the above system the shortest method is theelimination method which allow us to reduce the above system to a linear differential

equation of order n with constant coefficients. We will describe this method in the fol-

lowing two relevant examples. In practice when we have systems with three unknowns

we use the notations for the unknown functions: y = y(x), z = z(x), w = w(x), x Ror x = x(t), y = y(t), z = z(t), t R.

26

27

Example 3.1.1. Integrate the following systemy = y z + wz = y + z ww = 2y z

.

Solution. We will transform by elimination, the above system into a linear differ-

ential equation of order 3 in y. We write the first equation y y = z + w and wedifferentiate to obtain

y y = z + w (1).We replace y, z and w from the first, the second and the third equation of the systemto obtain

y (y z + w) = (y + z w) + 2y z.All the terms with y are kept in the left part hence

y 2y = 3z + 2w (2).

Again we differentiate and make the same replacements to obtain y2y = 3z+2whence y 2y+ 2z 2w = 3y 3z + 3w+ 4y 2z. We have a third equation withy in the left part

y 3y = 7z + 5w (3).We write the equivalent system formed by the equations (1),(2),(3) (viewed as a

system with 3 equations and two unknowns z, w) and we compute the rank of the

attached matrix z + w = y y3z + 2w = y 2y7z + 5w = y 3y

.

The attached matrix is

A =

1 13 27 5

which has rank 2. For the above system to be compatible the extended attached

matrix must have rank two as well, hence its determinant of order 3 must be zero.

28

We obtain the equation 1 1 y y3 2 y 2y7 5 y 3y

= 0that is

2y + 6y 15y + 15y 7y + 14y + 14y 14y 14y + 5y 10y + 3y 9y = 0.

We have y2yy+2y = 0 with the characteristic equation r32r2r+2 = 0 anddecomposing (r 2)(r2 1) = 0 we obtain the simple roots r1 = 1, r2 = 2, r3 = 1.The solution is y = C1e

x +C2e2x +C3e

x. For finding z we multiply equation (1) by2 and add equation (2) to get 2z3z = z = 2y+2y+y2y hence z = 2yyWe obtain

z = 2(C1ex + 2C2e

2x C3ex) (C1ex + 4C2e2x + C3ex)

= C1ex 3C3ex.

Finally from equation (1) we have

w = y y + z = C1ex + 2C2e2x C3ex C1ex C2e2x C3ex + C1ex 3C3ex

= C1ex + C2e

2x 5C3ex.

Example 3.1.2. Integrate the systemy = z + wz = w

w = y + w.

Solution. The first equation remains

y = z + w (1).

Differentiating we obtain y = z + w hence y = w y + w = y. The secondequation is now

y + y = 0 (2).

29

Differentiating this equation once more we get y+y = 0 and by replacing y = z+w(which is the first equation of the system) we obtain

y = z w (3).

Equations (1), (2) and (3) give us the system

z + w = y0 = y 2yz w = y

, with the matrix

A =

1 10 0

1 1

.This matrix has rank 1 and the extended matrix is

A =

1 1 y0 0 y + y

1 1 y

.We want that this matrix A to have the same rank 1 thus we require that all minors

of order 2 to be 0. We obtain 1 y0 y + y = 0

1 y1 y = 0,

which give us the resolvent equations

{y + y = 0

y + y = 0, with the characteristic equa-

tions

{r2 + 1 = 0

r3 + r = 0. We search for the common solutions ! Since the first equation

has roots r1,2 = i and the second equation has roots r1,2 = i, r3 = 0 we obtainthat

y = C1 cosx+ C2 sinx.

30

For finding w we use the third equation of the given system which tells us ww = yhence w w = C1 cosx C2 sinx. This is a linear non-homogenous equation oforder one with the homogenous solution wo = C3e

x. Since i is not a root for the

characteristic equation (r 1 = 0) we search for a particular solution of the formwp = A cosx+B sinx. We replace it to obtain

A sinx+B cosx A cosxB sinx = C1 cosx C2 sinx,

hence A + B = C1 and A B = C2 . We add these equation to get 2A =C1 C2 and we get A = C1+C22 , B = C2C12 . The solution w is now

w = wo + wp = C3ex +

C1 + C22

cosx+C2 C1

2sinx.

For finding z we use the second equation of the given system z = w, that isz =

(C3e

x + C1+C22

cosx+ C2C12

sinx)dx hence

w = C3ex +

C1 + C22

sinx C2 C12

cosx.

Example 3.1.3. Integrate the system x = x+ y + et

y = x+ y et, x = x(t), y = y(t), t R.

Solution. In the first equation we keep all terms in x on left side: x x = y+ et.We differentiate and replace x and y:

x x = y + et,

x x y et = x+ y et + et.We get x 2x = 2y + et. Form the first equation we have

y = x x et (1)

which we replace in the previous equation ,hence

x 2x = 2x 2x 2et + et

31

We obtain the linear non-homogenous equation of order 2

x 2x = et.

The characteristic equation is r2 2r = 0 with the roots r1 = 0, r2 = 2 hencexo = C1 + C2e

2t. Since 1 is not a root of the characteristic equation we search for a

particular solution of the form xp = Aet. We obtain Aet 2Aet = et hence A = 1

and the solution is

x = xo + xp = C1 + C2e2t + et.

For finding y we use (1) and we get

y = (C1 + C2e2t + et) (C1 + C2e2t + et) et = C1 + C2e2t et.

3.2 Symmetric Systems

A symmetric system of order n (where n is a non-negative integer) is a system of the

formdy1

f1(y1, . . . yn+1)= . . . =

dynfn(y1, y2, . . . , yn+1)

=dyn+1

fn+1(y1, . . . , yn+1),

where y1, . . . , yn+1 are the unknown functions. For solving such a system we search

for n independent prime integrals. A prime integral F is a constant map on any

solution of the system, that is a map of the form

F : Rn+1 R, F (y1, . . . , yn+1) = C.

In this book we concentrate on systems with 3 unknowns denoted

dx

f(x, y, z)=

dy

g(x, y, z)=

dz

h(x, y, z).

For this system we must find 3 independent prime integrals. We will solve also systems

with 4 unknowns denoted

dx

f(x, y, z, u)=

dy

g(x, y, z, u)=

dz

h(x, y, z, u)=

du

j(x, y, z, u).

For this kind of system we must find 3 independent prime integrals. To determine

prime integrals we have the following methods:

32

-if two rapports depend only on 2 unknowns, the equality of these 2 rapports is a

differential equation which in general can be integrated;

-if from a prime integral we can express an unknown as a function of the rest of

the unknowns we can reach sometimes to the above situation;

-we apply integrable combinations of the form

dy1f1

= . . . =dyn+1fn+1

=g1dy1 + . . . gn+1dyn+1g1f1 + . . .+ gn+1fn+1)

,

where we choose convenable functions (usually constant functions) g1, . . . gn+1 such

that

g1dy1 + . . . gn+1dyn+1 = dG

g1f1 + . . . gn+1fn+1 = f G.Next, we apply the above methods to this new system, and if necessary again the

above methods until we find n independent prime integrals. .

Example 3.2.1. Integrate the symmetric system

dx

z2 y2 =dy

z=

dz

y .

Solution. For the first prime integral we take the equality of the last two rapportsdyz

= dzy which become ydy = zdz. We integrate ydy = zdz to obtain y2

2=

z2

2+ C1, hence the first prime integral

y2 + z2 = C1.

Next we do some integrable combinations by amplifying the second rapport with

z, the third rapport with z and adding the obtained terms:

dx

z2 y2 =dy

z=

dz

y =zdy + ydz

z2 y2 .

We separate the equality of the first and the last rapport and using the well-known

formula d(yz) = dyz + ydz we obtain

dx

z2 y2 =d(yz)

z2 y2 ,

33

hence dx = d(yz) which is integrable with the solution x yz = C2. The solution ofthis system is {

y2 + z2 = C1

x yz = C2.

Example 3.2.2. Integrate the system y = y(y + z)

z = z(y + z).

Solution. This is a symmetric system since if we use y = dydx, z = dz

dxwe have

dy

y(y + z)=

dz

z(y + z)= dx.

We apply the first method for the first two rapports

dy

y(y + z)=

dz

z(y + z).

Simplifying with y + z we get the integrable equation dyy

= dzz

, which leads tody

y=

dz

z

with the prime integraly

z= C1.

Next we apply the second method by replacing y = zC1 and obtaining the equation

dz

z2(C1 + 1)= dx.

Integrating

dzz2(C1+1)

=dx we obtain the solution

1z(C1 + 1)

= x C2

hence the second prime integral (by replacing yz

= C1) is

1z(y

z+ 1)

= x C2,

that is

x+1

y + z= C2.

These two prime integrals are independent.

34

3.3 Partial differential equations

A homogenous partial differential equation of order one is an equation of the form

f1u

x1+ . . .+ fn

u

xn= 0

where f1 = f1(x1, . . . , xn), . . . , fn = fn(x1, . . . , xn) are functions depending on n vari-

ables x1, . . . , xn and u C2(Rn,R) is the unknown function. To solve this equationwe consider the symmetric system

dx1f1

= . . . =dxnfn

and we find n 1 prime integrals F1, . . . , Fn1. Now, for any function G on n 1variables the composed function

u = G(F1, . . . , Fn1)

is the general solution of the above partial differential equation.

A Cauchy problem for the equation

f1u

x1+ . . .+ fn

u

xn= 0,

is the problem of finding that solution for this equation which for a fixed value of

some variable, for example xi = a R, i {1, . . . , n}, we can reduce it (the solution)to a given function

y(x1, . . . , xi1, a, xi+1, . . . , xn) = g(x1, . . . , xi1, xi+1, . . . , xn).

We also say that we look for the integral surface which contains a given curve.

Example 3.3.1. Find the solution of the Cauchys problem for the following equation

xu

x+ y

u

y+ xy

u

z= 0

u(x, y, 0) = x2 + y2.

35

Solution. The symmetric system is dxx

= dyy

= dzxy

. We take the firs equality and

we integrate

dxx

=

dyy

to obtain ln |x| = ln |y|+ ln |C1|. The first prime integral isx

y= C1.

Next we amplify the first rapport with y, the second with x and we add to obtain

dx

x=dy

y=dz

xy=ydx+ xdy

2xy.

We separate the last equality to get dzxy

= d(xy)2xy

, hence we can integrate

2dz =d(xy)

to obtain xy 2z = C2. The general solution isu(x, y, z) = G(

x

y, xy 2z).

For the Cauchy problem we write the system formed by the prime integrals and the

initial conditions

xy

= C1

xy 2z = C2z = 0

u = x2 + y2

.

So, if we solve the system of the first 3 equations we get x = yC1 and xy = C2.

We multiply these 2 relations to obtain x2y = yC1C2 hence x2 = C1C2. Similarly

we obtain (by dividing the two relations) y2 = C2C1

, hence if we replace in the last

equation of the above system we get u = C1C2 +C2C1

. We use the prime integrals to

get the solution of Cauchys problem

u(x, y, z) =x

y(xy 2z) + xy 2zx

y

.

A quasi-linear partial differential equation of order one is an equation of the form

f1u

x1+ . . .+ fn

u

xn= g

where f1 = f1(x1, . . . , xn, u), . . . , fn = fn(x1, . . . , xn, u), g = g(x1, . . . , xn, u) are func-

tions depending on n + 1 variables x1, . . . , xn, u and u C2(Rn,R) is the unknownfunction. To solve this equation we consider the symmetric system

dx1f1

= . . . =dxnfn

=du

g

36

and we find n prime integrals F1, . . . , Fn. Notice that in this case the prime integrals

are depending on n+ 1 variables

F1 = F1(x1, . . . , xn, u), . . . , Fn = Fn(x1, . . . , xn, u).

Now, for any function G on n variables the relation

G(F1, . . . , Fn) = 0

is the implicit solution of the above partial differential equation. As in the case of

homogenous equation we can solve similarly Cauchys problems.

Example 3.3.2. Find the general solution of the equation

xuu

x+ yu

u

y= xy

and the surface passing through the curve y = x2, u = x3.

Solution. The symmetric system is

dx

xu=dy

yu=

du

xy .

The equality of the first rapports gives us the prime integral dxxu

= dyyu

, that is

x

y= C1 (1.)

Next we amplify the first rapport with y, the second with x and we add to obtain

dx

xu=dy

yu=

du

xy =ydx+ xdy

2xyu.

The equality of the last rapports is duxy =ydx+xdy2xyu

, thus

du

xy =d(xy)

2xyu.

We integrate

2udu = d(xy) to obtainu2 + xy = C2 (2).

37

The general solution is

G(x

y, u2 + xy) = 0.

To solve the Cauchy problem, or equivalently to find the surface passing through the

given curve we take x as a parameter:

x = x, y = x2, u = x3

and we replace these relations in the prime integrals (1) and (2) to obtain

1

x= C1, x

6 + x3 = C2.

We eliminate x in these equations to get

1

C61+

1

C31= C2.

Using again the prime integrals (1) and (2) we obtain the final solution(yx

)6+(yx

)3= u2 + xy.

Example 3.3.3. Find the solution of the following Cauchy problem

xu

x+ (xu+ y)

u

y= u

satisfying the initial conditions x+ y = 2u, xu = 1.

Solution. The symmetric system is

dx

x=

dy

xu+ y=du

u.

The equality of the first and the third rapport dxx

= duu

gives by integration

x

u= C1.

From this prime integral we express x as x = C1u and we replace it in the symmetric

systemC1du

C1u=

dy

C1u2 + y=du

u.

38

Next we have dydu

= C1u2+yu

hence we have the equation

dy

du= C1u+

y

u(1).

We denote yu

= t so dy = tdu+ udt. It follows that

dy

du= t+ u

dt

du(2).

From (1) and (2) we have t+u dtdu

= C1u+ t so udtdu

= C1u. We get dt = C1du, that is

t = C1u+C2. Moreover sinceyu

= t and xu

= C1 we obtain the second prime integral

y

u x = C2.

The general solution is

G(x

u,y

u x) = 0.

For the Cauchy problem we take, in the initial conditions, u as parameter, that is

x =1

u, y = 2u 1

u, u = u.

We replace these in the prime integrals for obtaining

1

u2= C1 (3)

2u 1u

u 1u

= C2 (4).

We eliminate u, by using u = 1C1

, obtained from (3) and replacing in (4)

2C1C11C1

C1 = C2.

We have that 2C1C1 = C2 and replacing the above prime integrals we find the

final solution

2 xux

u=y

u x.

39

3.4 Problems

Problem 3.4.1. Integrate the following linear system with constant coefficients

(1).

y = 2y + z

z = 3y + 4z;

(2).

y = y zz = z 4y

;

(3).

y = y z + wz = y + z ww = 2y z

;

(4).

y = y z wz = y + z

w = 3y + w

;

(5).

y = 3y z + wz = y + z + w

w = 4y z + 4w;

(6).

y = y z + wz = y + z ww = 2w z

;

(7).

x = y + 2et

y = x+ t2;

40

(8).

x = y 5 cos ty = 2x+ y

;

(9).

x = 3x+ 2y + 4e5t

y = x+ 2y;

(10).

x = 2x 4y + 4e2t

y = 2x 2y;

(11).

x = 2x+ y 2z ty = 1 xz = x+ y z t

;

(12)

x = 4x+ y e2t

y = y 2x.

Problem 3.4.2. Integrate the symmetric systems

(1). dxy+z

= dyx+z

= dzy+x

;

(2). dxyz

= dyxz

= dzyx

;

(3). dxz2y2 =

dyz

= dzy ;

(4). dxx(y+z)

= dyz(zy) =

dzy(yz) .

Problem 3.4.3. Find the general solution for the following partial differential equa-

tions

(1). y ux xu

y= 0;

(2). xux

+ y uy

+ z uz

= 0;

41

(3). y ux

+ xuy

= x y;

(4). 2xux

+ (y x)uy x2 = 0;

(5). xux

+ y uy

+ (z + u)uz

= xy.

Problem 3.4.4. Find the general solution and the solution for the Cauchy problem

of the following equations

(1). xux y u

y= 0, u(x, 1) = 2x;

(2). 2xux y u

y= 0, u(1, y) = y2;

(3). ux

+ uy

+ 2uz

= 0, u(1, y, z) = yz;

(4). xux

+ y uy

+ xy uz

= 0, u(x, y, 0) = x2 + y2;

(5). xux 2y u

y= x2 + y2, y = 1, u = x2;

(6). xux

+ y uy

= u x2 y2, y = 2, u = x x2;

(7). uux

+ (u2 x2)uy

+ x = 0, y = x2, u = 2x;

(8). xux

+ (xu+ y)uy

= u, x+ y = 2u, xu = 1.

Solutions: 3.4.1 (1). y = C1ex + C2e

5x, z = C1ex + 3C2e5x; (2). y = C1ex +C2e

3x, z = 2C1ex2C2e3x; (3). y = C1ex+C2e2x+C3ex, z = C1ex3C3ex, w =

C1ex + C2e

2x 5C3ex; (4).y = ex(2C2 sin 2x + 2C3 cos 2x), z = ex(C1 C2 cos 2x +C3 sin 2x), w = e

x(C1 3C2 cos 2x+ 3C3 sin 2x); (5).y = C1ex +C2e2x +C3e5x, z =C1e

x2C2e2x+C3e5x, w = C1ex3C2e2x+3C3e5x; (6). y = (C1+C2x)ex+C3e2x, z =(C12C2 +C2x)ex, w = (C1C2 +C2x)ex+C3e2x; (7). x = C1et+C2et+ tet t22, y = C1e

t C2et + (t 1)et 2t; (8).x = C1e2t +C2et + tet 2 sin t cos t, y =2C1e

2tC2et + sin t 3 cos t; (9). x = C1et + 2C2e4t + 3e5t, y = C1et +C2e4t + e5t;(10). x = C1(cos 2t sin 2t) + C2(cos 2t + sin 2t), y = C1 cos 2t + C2 sin 2t + e2t;(11). x = C1e

t +C2 cos t+C3 sin t, y = C1etC2 sin tC3 cos t+ 1, z = C2 cos t+

42

C3 sin t t12 ; (12). x = C1e2t +C2e3t + (t+ 1)e2t, y = 2C1e2tC2e3t 2te2t. 3.4.2(1). yx

zx = C1, (x+y+z)(zx)2 = C2; (2). y2x2 = C1, z2x2 = C2; (3). y2+z2 =C1, yz x = C2; (4). y2 + z2 = C1, x(y z) = C2. 3.4.3 (1). u(x, y) = G(x2 + y2);(2). u(x, y, z) = G( y

x, zx); (3). G(x2 y2, x y + u) = 0; (4). G(x2 4u, (x+y)2

x) = 0;

(5). G(xy, xy 2u, z+uxy

x) = 0. 3.4.4 (1). u(x, y) = 2xy; (2). u(x, y) = y2e2

x2;

(3). u(x, y, z) = (1 x + y)(2 2x + z); (4). u(x, y, z) = (xy 2z)(xy

+ yx); (5).

2x2(y + 1) = y2 + 4u 1; (6). x 2y = x2 + y2 + u; (7). x2 + u2 = 5(xu y); (8).xu = (xu y x+ 2u)2.

Chapter 4

Complex Analysis

4.1 Complex numbers. Basic results

Recall that the set of complex numbers is denoted C. A complex number z Chas an algebraic form z = x + iy where x = Rez R is the real part of z andy = Imz R is the imaginary part of z. The symbol i is the imaginary unity andhas the fundamental property that i2 = 1. The conjugate of a complex number z isdenoted z = x iy.

In some exercises is important to remember the following formulas

Rez = x =z + z

2;

Imz = y =z z

2i.

We recall some basic properties of complex numbers in the next proposition. The

proof is proposed as a problem in the last section.

Proposition 4.1.1. Let z1, z2 C. The next statements holds

a) z1 z2 = z1 z2;

b) z1z2 = z1z2;

43

44

c)(z1z2

)= z1

z2

We associate to z its geometrical image M(x, y) in the plane xOy. The module of

a complex number z is the non-negative real number |z| = x2 + y2 and representsthe distance form the origin O to M . A complex number has also a trigonometrical

form

z = |z|(cos + i sin )where = argz is the angle between Ox and OM in a positive sense and is called the

principal argument of z. The set of all arguments is denoted

Argz = {argz + 2kpi | k Z}.

To obtain the principal argument we apply argz = arctan yx

+ kpi with k = 0 if M is

in the first quadrant, k = 1 if M is in the second or third quadrant and k = 2 if M

is in the fourth quadrant. To recall the operations of complex numbers we solve the

following problem.

Problem 4.1.2. Compute the following expressions:

a) (2 + i) 5(1 + i) + 2 + 3i;

b) (1 4i)(3 + i);

c) i+15+2i

;

Solution.

a) (2+ i)5(1+ i)+2+3i = 2+ i55i+2+3i = 25+2+ i5i+3i = 1 i;

b) (1 4i)(3 + i) = 3 i 12i 4i2 = 3 4(1) i 12i = 1 13i;

c) To compute fractions of complex numbers (to divide complex numbers) we

amplify the fraction with the conjugate of the denominator

i+ 1

5 + 2i=

(i+ 1)(5 2i)(5 + 2i)(5 2i) =

5i 2i2 + 5 2i52 4i2 =

7 + 3i

29.

45

For multiplication it is sometimes more convenient to use the trigonometrical form.

So let z1 = |z|(cos 1 + i sin 1), z2 = |z2|(cos 2 + i sin 2) and let n be a nonnegativeinteger. Then

z1z2 = |z1| |z2|[cos(1 + 2) + i sin(1 + 2)];zn = |z|n[cos(n) + i sin(n)].

The roots of order n are given using the trigonometrical form:

zk =n|z|(

cos + 2kpi

n+ i sin

+ 2kpi

n

)are the n distinct roots of z where k = 0, . . . , n 1.

We identify sets of complex numbers with sets of points from the complex plane.

For example the circle with center z0 and radius r > 0 is

C(z0; r) = {z C; | z z0| = r}.

The disc with center z0 and radius r > 0 is

D(z0; r) = {z C; |z z0| < r},

and the circular crown with center z0 and radiuses r, R is

C(z0; r;R) = {z C; r < |z z0| < R}.

Let (zn)n0 be a sequence of complex numbers where zn = xn + iyn or

zn = n(cos n + i sin n), n 0, n [0, 2pi), n 0.

As in the real case we say that the sequence is convergent if there is z C such thatlimn

zn = z. We have that

limn

zn = z = x+ iy

if and only if limn xn = x, limn yn = y. Similarly we can prove that

limn

zn = (cos + i sin ), [0, 2pi)

if and only if limn

|zn| = |z| = and if 6= 0, limn

arg(zn) = argz = . Using this

last formula if we denote with ez the limit limn

(1 +

z

n

)nwe obtain Eulers formula

ez = ex+iy = ex(cos y + i sin y).

46

4.2 Complex functions of a complex variable

A complex function of a complex variable has the form

f : D C, f(z) = u(x, y) + iv(x, y)where u = Ref, v = Imf and z = x + iy, x, y R, D C. Let z0 = x0 + iy0 be acluster point of D. The derivative of f at z0 is defined by

f (z0) = limzz0

f(z) f(z0)z z0 ,

if the previous limit exists. If f (z0) is finite we say that f is differentiable at z0. Afunction which is differentiable at every point of a domain D is called holomorphic

on D.

Theorem 4.2.1. If f is differentiable at z0 then the following relations are satisfied

(C R)

ux

(x0, y0) =vy

(x0, y0)

uy

(x0, y0) = vx(x0, y0)

(C-R) are called the Cauchy-Riemann equations. The converse is also true in

appropriate conditions.

Theorem 4.2.2. If u and v are function of class C1 in a neighborhood of z0 and the

conditions (C-R) are satisfied then f is differentiable at z0. Moreover

f (z0) =u

x(x0, y0) + i

v

x(x0, y0).

In then next lines we recall some properties of holomorphic functions. We denote

by H(D) the set of all functions f : D C which are holomorphic on D. Letf H(D), f = u+ iv, u, v : D R.

Proposition 4.2.3. If u or v are constant functions on D then f is a constant

function on D.

47

Proposition 4.2.4. The real and the imaginary part of the holomorphic function f

are harmonic functions on D, i.e.

u =2u

x2+2u

y2= 0

v =2v

x2+2v

y2= 0,

on D.

Proposition 4.2.5. If the real part of the holomorphic function f is known then f

is determined up to a constant. The imaginary part v can be determined from

v(x, y) = xx0

u

y(t, y)dt+

yy0

u

x(x0, t) + k, k R,

where (x0, y0) D and D is a simple connected domain.Some complex functions. We define the exponential, sinus and cosinus by the

folllowing power series

ez = 1 +z

1!+z2

2!+z3

3!+ . . . ,

cos z = 1 z2

2!+z4

4!+ . . . ,

sin z = z z3

3!+z5

5! . . . ,

for any z C.Eulers relation is satisfied

eiz = cos z + i sin z, z C.

For z C \ {0} the complex logarithm is defined by

Logz = ln |z|+ i(argz + 2kpi), k Z.

For k = 0 we get ln z = ln |z| + iargz is called the principal branch of Logz. Thepower function is defined by

z = eLogz, z C \ {0}, C.

48

From Eulers formula we get

cos z =eiz + eiz

2, sin z =

eiz eiz2i

, z C.

The hyperbolic functions are

chz =ez + ez

2, shz =

ez ez2

, z C.

The following relations hold

sin2 z + cos2 z = 1; ch2z sh2z = 1, z C.

We continue this section with some important solved problems.

Problem 4.2.6. Find the real part and the imaginary part of the following numbers:

a) sin(2 i);

b) ii;

Solution.

a)

sin(2 i) = ei(2i) ei(2i)

2i=e1+2i e12i

2i

=e(cos 2 + i sin 2) e1(cos 2 + i sin 2)

2i=

(e 1e) cos 2 + i(e+ 1

e) sin 2

2i

=i(e 1

e) cos 2 (e+ 1

e) sin 2

2 =e2 + 1

2esin 2 + i

1 e22e

cos 2.

b) ii = eiLogi, where

Logi = {ln |i|+ i(argi+ 2kpi) | k Z}.

Since |i| = 12 + 02 = 1 and argi = arctan 10

= arctan = pi2

we obtain that

Logi = {i(pi2

+ 2kpi) | k Z},

thus

ii = {ei(pi2+2kpi) | k Z}.

49

Problem 4.2.7. Find the holomorphic functions f : C C, f(z) = u(x, y) +iv(x, y), x, y R if:

a) u(x, y) = x2 y2 x;

b) v(x, y) = yx2+y2

, f(1) = 0;

Solutions.

a) Solution 1. From the (C-R) equations we get ux

= vy, u

y= v

x, hence the

relationsv

y= 2x 1; (4.2.1)v

x= 2y; (4.2.2)

From (4.2.1) and (4.2.2) it follows

dv = 2ydx+ (2x 1)dy = P (x, y)dx+Q(x, y)dy,

thus

v(x, y) =

x0

P (t, y)dt+

y0

Q(0, t)dt+ k =

x0

2ydt+

y0

(1)dt

= 2xy y + k, k R.Now the expression of f is

f(z) = u(x, y) + iv(x, y) = x2 y2 x+ i(2xy y) + ik

= (x2 y2 + 2ixy) (x+ iy) + ik = z2 z + ik.

Solution 2. Integrate the above relation (4.2.1) to obtain v(x, y) = (2x 1)y +(x). Now we apply (4.2.2) and it follows that 2y + (x) = 2y, thus (x) = 0and (x) = k, k R. It follows analogously f(z) = z2 z + ik, k R.

b) We haveu

x=v

y=x2 + y2 2y2

(x2 + y2)2; (4.2.3)

y

y= v

x=

2xy

(x2 + y2)2; (4.2.4)

50

From (4.2.4) we have that u(x, y) =

2xy(x2+y2)

dy = xx2+y2

+ (y). Replacing in

(4.2.3) we get ux

= x2y2

(x2+y2)2hence

x2 + y2 2x2(x2 + y2)2

+ (y) =x2 y2

(x2 + y2)2.

We obtain (y) = 0, that is (y) = k, k R. We get u(x, y) = xx2+y2

+ k.

f(z) = xx2 + y2

+ k + iy

x2 + y2=x+ iyx2 + y2

+ k

= x iy(x+ iy)(x iy) + k

= 1x+ iy

+ k =1z

+ k.

The condition f(1) = 0 leads to k = 1, so

f(z) = 1 1z, z C \ {0}.

Problem 4.2.8. Find all holomorphic functions f(z) = u(x, y) + iv(x, y), z = x +

iy, x, y R if u(x, y) = (x2 y2), C2(R).

Solution. The function u satisfies Laplace equation u = 0, that is 2ux2

+ 2uy2

= 0.

We obtainu

x= (x2 y2)2x;

2u

x2= 2(x2 y2) + 4x2(x2 y2);u

y= (x2 y2)(2y);

2u

y2= 2(x2 y2) + 4y2(x2 y2).

Since u = 0 it follows that for any (x, y) R2 we have 4(x2 + y2)(x2 y2) = 0hence (x2 y2) = 0. Let x2 y2 = t, t R. The relation (t) = 0 implies

(t) = C1t+ C2, C1, C2 R,

51

so u(x, y) = C1(x2 y2) + C2. (C-R) equations lead to

v

x= 2C1y

v

y= 2C1x.

The first relation leads to v(x, y) =

2C1ydx = 2C1xy + (y). Replacing in the

second relation we obtain 2C1 + (y) = 2C1x hence (y) = 0 and (y) = k, k R

so v(x, y) = 2C1xy + k. Finally

f(z) = C1(x2 y2) + C2 + i(2C1xy + k)

= C1(x2 y2 + 2ixy) + C2 + ik= C1z

2 + C2 + ik.

Let = C2 + ik, C then f(z) = Cz2 + ,C R.

Problem 4.2.9. Find z C such that sin z = 4i3,

Solution. The equation is equivalent to eizeiz

2i= 4i

3, that is eiz eiz = 8

3. We

obtain the equation e2iz + 83eiz 1. Let eiz = w. Then 3w2 + 8w 3 = 0 which is an

equation with = 100 and w1 =13;w2 = 3.

In the first case we obtain eiz = 13

with

iz = Log1

3= ln |1

3|+ i(arg1

3+ 2kpi) = ln 3 + 2kpii.

In the second case we obtain eiz = 3 with the solution

iz = Log(3) = ln | 3|+ i(arg(3) + 2kpi) = ln 3 + (2k + 1)pii.

The solution is

z {2kpi + i ln 3|k Z} {(2k + 1)pi i ln 3|k Z}.

Problem 4.2.10. Prove the following relations:

a) Re sin z = sinxchy;

52

b) | sin z| =

ch2y cos2 x.

Solution.

a)

sin z =eiz eiz

2i=ei(x+iy) ei(x+iy)

2i=eixeiy eixey

2i

=(cosx+ i sinx)ey (cosx i sinx)ey

2i

=cosxey + i sinxey cosxey + i sinxey

2i

=cosx(ey ey) + i sinx(ey + ey)

2i=

sinx(ey + ey)2

+ icosx(ey ey)

2

= sinxchy + i cosxshy;

b) The above relation sin z = sinxchy + i cosxshy leads to

| sin z| =

sin2 xch2y + cos2 xsh2y.

It is well known that sin2 x = 1 cos2 x, sh2y = ch2y 1 hence

| sin z| =

(1 cos2 x)ch2y + cos2 x(ch2y 1) =

ch2y cos2 x.

4.3 Laplaces transform.

Laplaces transform is defined for any function f : R C by

L{f(t)} = F (s) = 0

estf(t)dt,

where s D C. By D we denote the set of complex numbers s for which the aboveintegral is convergent. If D 6= we say that the function f has Laplaces transformon D.

A function f : R C is called an original if the following statements are true

f(t) = 0, t < 0;

f has a finite number of first septets? discontinuity for any bounded interval;

53

f has an exponential growing order, that is there are constants M, 0 suchthat |f(t)| Met,t > 0

The set of all original functions is denoted by O. The constant is called a

growing index and the smallest growing index is denoted o = o(f).

Theorem 4.3.1. If f O then F (s) exists and is holomorphic in the semiplaneRes > o(f).

In the next example we show how to compute the Laplaces transform using its

definition.

Example 4.3.2. For any a C we have

L{eat} = 0

eatestdt =[e(as)t

a s]0

=1

s a,

for all s with Res > Rea.

However in practice we will use the properties of Laplaces transform which we will

give in the next theorem and a basic list of Laplaces transform for some elementary

functions.

Theorem 4.3.3. (1) (Linearity). For any , C and f, g O we have

L{f(t) + g(t)} = L{f(t)}+ L{g(t)};

(2) If f O and a > 0 then L{f(at)} = 1aF ( s

a);

(3) If f O and a CC then L{eatf(t)} = F (s a) for Re s > Re a+ o(f);

(4) If f O and a > 0 then L{f(t a)} = easF (s) for Re s > o(f);

(5) (The differentiation of the original). If f is continuous and f, f , f O thenfor any s such that Re s > max{o(f), o(f )} we have

L{f (t)} = sF (s) f(0 + 0),

54

and for any s such that Re s > max{o(f), o(f ), o(f )} we have

L{f (t)} = s2F (s) sf(0 + 0) f (0 + 0);

(6) (The differentiation of the transform). If f O is a continuous function thenfor any s such that Re s > 0(f) we have

F (s) = L{tf(t)}.

(7) (The integration of the transform). If f(t)t O then for any s such that Re s >

o(f(t)t

) we have

L

{f(t)

t

}=

s

F (s)ds;

In particular 0

f(t)

t=

0

F (s)ds.

It is recommended that the next list which contains some fundamental Laplaces

transforms of elementary functions to be memorized.

Table with Laplaces transforms:

(1) L{eat} = 1sa for a C, Res > Re a;

More generally for n 0 any integer, we have

L{eattn} = n!(s a)n+1 ;

In particular L{1} = 1s

and L{at} = as2

;

(2) L{sin at} = as2+a2

for a > 0;

(3) L{cos at} = ss2+a2

for a > 0;

(4) L{sh at} = as2a2 for a > 0;

(5) L{ch at} = ss2a2 for a > 0;

55

To see how to apply the above formulas we present some solved problems.

Problem 4.3.4. Compute the transform L{cos2 t}.Solution. The well-known trigonometrical formula cos2 t = 1+cos 2t

2gives us

L{cos2 t} = 12

(L{1}+ L{cos 2t}) = 12

(1

s+

s

s2 + 22

)=

1

2 2s

2 + 4

s3 + 4s.

For the second equality we used formula (1) and (3) from the above table.

Problem 4.3.5. Determine the next improper integral0

cos 2tcos tt

dt.

Solution. From Theorem 4.3.3,(7) we have that0

cos 2tcos tt

dt =0F (s)ds,

where F (s) = L{cos 2t cos t}. ThenF (s) = L{cos 2t} L{cos t} = s

s2 + 4 ss2 + 1

.

The above integral is now 0

cos 2t cos tt

dt =

0

s

s2 + 4ds

0

s

s2 + 1ds

=1

2

[lns2 + 4

s2 + 1

]0

=1

2(0 ln 4) = ln 1

2.

Laplaces transform has also an inverse. To compute the inverse of Laplace s

transform we have the next theorem.

Theorem 4.3.6. (Mellin-Fourier.)

If f O and L{f(t)} = F (s), Res > o(f) then

f(t) = L1{F (s)} = 12ipi

x+ixi

estF (s)ds, x > o(f).

The properties from Theorem 4.3.3 can be translated to its inverse. If we know

the table for the direct transform then we can find easily the inverse. For example:

(I1). L1{ 1(sa)n+1} = e

attn

n!;

(I2). L1{ as2+a2

} = sin at;(I3). L1{ s

s2+a2} = cos at.

Using Laplaces transform and its inverses we can easily solve Cauchys problems

for linear equations or linear system with constant coefficients

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Problem 4.3.7. Solve Cauchys problem for the differential equation

y 4y = e2t + e2t

2, y(0) = y(0) = 0.

Solution. We denote by Y the Laplaces transform

L{y(t)} = Y (s).

The differentiation formulas Theorem 4.3.3,(6) and the initial conditions of Cauchys

problem give us

L{y(t)} = sY (s) y(0 + 0) = sY (s);L{y(t)} = s2Y (s) sy(0 + 0) y(0 + 0) = s2Y (s).

We apply Laplaces transform on the given differential equation to obtain the

equation in Y :

s2Y (s) 4Y (s) = 12L{e2t}+ 1

2L{e2t},

hence

(s2 4)Y (s) = 12

(1

s 2 +1

s+ 2

)=

1

2 s+ 2 + s 2

s2 4 =s

s2 4 .

It follows that Y (s) = s(s24)2 and we want to find A,B,C,D constants such that

Y (s) =s

(s 2)(s+ 2)2 =As+B

(s 2)2 +Cs+D

(s+ 2)2.

We obtain s = (As+B)(s+ 2)2 + (Cs+D)(s 2)2, which give us the systemA+ C = 0

4A+B 4C +D = 04A+ 4B + 4C 4D = 14B + 4D = 0

The solutions are A = 0, C = 0, B = 18, D = 1

8, thus

Y (s) =1

8

[1

(s 2)2 1

(s+ 2)2

].

57

We apply the inverse L1 to obtain

y(t) =1

8L1

{1

(s 2)2} 1

8L1

{1

(s+ 2)2

}=

1

8e2tt 1

8e2tt =

t

8(e2t e2t).

For the second equality we used (I1).

Problem 4.3.8. Integrate the system x(t) = x(t) + 2y(t), x(0) = 1

y(t) = 2x(t) + y(t), y(0) = 1

Solution. Consider the notations L{x(t)} = X(s), L{Y (t)} = Y (s) and apply Lon the equations to obtain{

L{x(t)} = L{x(t)}+ 2L{y(t)}L{y(t)} = 2L{x(t)}+ L{y(t)}

hence {sX 1 = X + YsY + 1 = 2X + Y

that is, the linear system {(s 1)X 2Y = 12X + (s 1)Y = 1

.

To solve this system we multiply the first equation by 2, the second equation by s1and we add to get

4Y + (s 1)2Y = 2 s+ 1,hence

Y =3 s

4 + s2 2s+ 1 =3 s

s2 2s+ 3 =3 s

(s 3)(s+ 1) =1s+ 1

.

Then

y(t) = L1{

1

s+ 1

}= et.

Similarly we obtain X(t) = 1s+1

and then

x(t) = L1{

1

s+ 1

}= et.

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4.4 Problems

Problem 4.4.1. Let z1 = 2 + 11i, z2 = 2 i. Compute or find the algebraic formx+ iy, x, y R where is the case:

a) Imz1;

b) z1z2;

c) z1z2

;

d) Rez21 , (Rez1)2.

e) |z1|;

f) |z1z2|;

g) |z1||z2|;

h) |z1 + z2|;

i) |z1|+ |z2|; Compare the result with the result of h) and explain geometrically.

Problem 4.4.2. Prove the relations from Proposition 4.1.1 by using that z1 = x1 +

iy1, z2 = x2 + iy2, x1, y1, x2, y2 R.

Problem 4.4.3. Find the real part and the imaginary part of:

a) ch(2 + i);

b) 1i;

c) Log(1 + i);

59

d) cos z, where z = x+ iy C;

e) z2z, where z = x+ iy C.

Problem 4.4.4. Find the holomorphic functions

f : C C, f(z) = u(x, y) + iv(x, y), x, y R

if:

a) u(x, y) = x2 y2 + xy, f(0) = 0;

b) u(x, y) = ex(x cos y y sin y), f(0) = 0;

c) v(x, y) = arctan yx, x > 0, f(1) = 0;

d) u(x, y) = x2 y2 + 2x, f(i) = 2i 1;

Problem 4.4.5. Find z C such that:

a) shz = i2;

b) chz = i2;

c) cos z = 5;

d) sin z = 0.

Problem 4.4.6. Prove the following relations, for any z = x+ iy C, x, y R:

a) Re cos z = cosxchy;

b) | cos z| =

ch2y sin2 x;

c) cos z = cosxchy i sinxshy;

60

d) sin z = sinxchy + i cosxshy;

e) |Rez| |z|;

f) |Imz| |z|;

g) (Parallelogram equality)

|z1 + z2|2 + |z1 z2|2 = 2(|z1|2 + |z2|2),

where z1, z2 C. Explain the name!

Problem 4.4.7. Find Laplaces transforms L{f(t)} for the next functions:

(1) f(t) = sin2 t;

(2) f(t) = sin3 t;

(3) f(t) = cos3 t;

(4) f(t) = e2t sin t;

(5) f(t) = t sin3 t;

(6) f(t) = 1cos tt

;

(7) f(t) = et1t

;

Problem 4.4.8. Compute the integrals:

(1)0

cos 2tcos 3tt

dt;

(2)0

ete2tt

dt;

(3)0

sin tsin 2tt

dt;

61

Problem 4.4.9. Integrate the following linear differential equations and systems of

linear differential with constant coefficients:

(1) y + 4y = cos 2t, y(0) = y(0) = 0;

(2) y + y = sin t, y(0) = y(0) = 0;

(3) y 4y = ch2t, y(0) = y(0) = 0;

(4) y 9y = sh3t, y(0) = y(0) = 0;

(5)

x = y + 2et

y = x+ t2, x(0) = y(0) = 1

Solutions 4.4.1 a) 11; b) 7 + 24i; c) 3 + 4i; d) 117, 4; e) 125; f) 25; g) 25;h) 10; i) 6

5; This number is larger than the result from h), since it may be viewed

as the sum of two sides of a triangle which is always larger than the third side: the

triangle inequality. 4.4.3 a) 12e

[(e2 + 1) cos 1 + i(e2 1) sin 1]; b) e2kpi, k Z; c)ln

2 + i(pi2

+ 2kpi), k Z; d) cosxchy i sinxshy; e) x3 + xy2 + i(y3 + x2y); 4.4.4a) f(z) = 1 i

2z2; b) f(z) = zez; c) f(z) = ln z; d) f(z) = z2 + 2z; 4.4.5 a)

z {(pi6

+ 2kpi)i|k Z} {(5pi6

+ 2kpi)ipik Z}; c)z {2kpi + i ln(5 24)|k Z};d)z {kpi|k Z}; 4.4.6 See 4.2.10; 4.4.7 (1) 2

(s2+4)s; (2) 6

(s2+1)(s2+9), use sin3 t =(

eiteit2i

)3; (3) s(s

2+7)(s2+1)(s2+9)

; (4) 1(s2)2+1 ; (5)

24s(s2+5)(s2+1)2(s2+9)2

, use Theorem 4.3.3,(6); (6)

ln s+1s

, use Theorem 4.3.3,(7); (7) ln ss1 , use Theorem 4.3.3,(7). 4.4.8 (1) ln

32; (2)

ln 2; (3) ln3; 4.4.9 (1) y(t) = t4

sin 2t; (2) y(t) = 12(sin tt cos t); (3) y(t) = t

4sh2t;

(4) y(t) = 118

(3tch3t sh3t); (5) x(t) = 52et + 1

2et + tet t2 2, y(t) = 5

2et 1

2et +

(t 1)et 2t.

Bibliography

[1] E. Kreyszig, Advance Engineering Mathematics 10 edition, John Wiley and Sons,

Inc., Hoboken,New Jersey, 2011.

[2] A. Philippov, Recueil de Problemes Dequations Differentielles, Editions Mir,

Moscou, 1976.

[3] S. Toader and G. Toader, Matematici Speciale I,II, U.T. Press, Cluj-Napoca, 2009.

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